Replacing G by Embedding Scaling Factors into Units of Measurement

James Rogers, SE Ohio, 22 Nov 2024 2317 

Abstract

The gravitational constant $G$ is traditionally viewed as a fundamental constant governing the strength of gravitational interactions. However, by redefining the units of mass (kilogram) and length (meter), we demonstrate that $G$ can be understood as a unit-scaling factor. This perspective simplifies gravitational equations by incorporating these scaling factors directly into the units of measurement. This paper explores the implications of such a redefinition and how it can fundamentally alter our understanding and application of gravitational laws.

Introduction

The gravitational constant $GG$ appears in Newton's law of universal gravitation:

$$F=\frac{G{m}_1{m}_2}{r^2}F\; =\; \backslash frac\{G\; m\_1\; m\_2\}\{r^2\}$$

where $F$ is the gravitational force, $m_1$ and $m_2$are the masses, and $r$ is the distance between them. The value of $G$ depends on the units used for mass, length, and time. By scaling these units, we can eliminate $G$ from the equation, simplifying the relationship between the masses and distance.

Methodology

We begin by examining the relationship between the constants $G$ and $hc$. The constants are expressed as:

• $hc=1.986446\times 10^{-25}\text{kg}\cdot {\text{m}}^3\mathrm{/}{\text{s}}^{2}hc\; =\; 1.986446\; \backslash times\; 10^\{-25\}\; \backslash ,\; \backslash text\{kg\}\; \backslash cdot\; \backslash text\{m\}^3/\backslash text\{s\}^2$

• $G=6.674301\times 10^{-11}{\text{m}}^3\mathrm{/}(\text{kg}\cdot {\text{s}}^2)G\; =\; 6.674301\; \backslash times\; 10^\{-11\}\; \backslash ,\; \backslash text\{m\}^3/(\backslash text\{kg\}\; \backslash cdot\; \backslash text\{s\}^2)$

• Ratio $\frac{hc}{G}=2.976260\times 10^{-15}\backslash frac\{hc\}\{G\}\; =\; 2.976260\; \backslash times\; 10^\{-15\}$

By redefining the kilogram such that $hchc$ and $GG$ become numerically equal, we find the necessary scaling factor.

Calculation of Scaling Factor

To make the ratio $\frac{hc}{G}=1\backslash frac\{hc\}\{G\}\; =\; 1$:

$$\text{kg scaling factor}=\sqrt{\frac{hc}{G}}\approx 5.455511\times 10^{-8}\backslash text\{kg\; scaling\; factor\}\; =\; \backslash sqrt\{\backslash frac\{hc\}\{G\}\}\; \backslash approx\; 5.455511\; \backslash times\; 10^\{-8\}$$

After Scaling

• New $hh$: $1.214564\times 10^{-26}(\text{scaled\_kg})\cdot {\text{m}}^2\mathrm{/}\text{s}1.214564\; \backslash times\; 10^\{-26\}\; \backslash ,\; (\backslash text\{scaled\backslash \_kg\})\; \backslash cdot\; \backslash text\{m\}^2/\backslash text\{s\}$

• New $hchc$: $3.641173\times 10^{-18}(\text{scaled\_kg})\cdot {\text{m}}^3\mathrm{/}{\text{s}}^{2}3.641173\; \backslash times\; 10^\{-18\}\; \backslash ,\; (\backslash text\{scaled\backslash \_kg\})\; \backslash cdot\; \backslash text\{m\}^3/\backslash text\{s\}^2$

• New $GG$: $3.641173\times 10^{-18}{\text{m}}^3\mathrm{/}(\text{scaled\_kg}\cdot {\text{s}}^2)3.641173\; \backslash times\; 10^\{-18\}\; \backslash ,\; \backslash text\{m\}^3/(\backslash text\{scaled\backslash \_kg\}\; \backslash cdot\; \backslash text\{s\}^2)$

Implications for mass:

• $1\text{kg}1\; \backslash ,\; \backslash text\{kg\}$ in original units = $5.455511\times 10^{-8}\text{kg}5.455511\; \backslash times\; 10^\{-8\}\; \backslash ,\; \backslash text\{kg\}$ in new units

• $1\text{kg}1\; \backslash ,\; \backslash text\{kg\}$ in new units = $1.833009\times 10^7\text{kg}1.833009\; \backslash times\; 10^7\; \backslash ,\; \backslash text\{kg\}$ in original units

Further Redefinition for Natural Units

To make both $hchc$ and $GG$ equal to 1 in their respective units, we redefine the meter.

Calculation of Scaling Factor for Meter

Scaling factor for meter:

\[ \left(3.641173 \times 10^{-18}\right){1/3} \approx 1.533474 \times 10^{-6} \]

New definitions:

• $1\text{new kg}\approx 1.833009\times 10^7\text{old kg}1\; \backslash ,\; \backslash text\{new\; kg\}\; \backslash approx\; 1.833009\; \backslash times\; 10^7\; \backslash ,\; \backslash text\{old\; kg\}$

• $1\text{new meter}\approx 1.533474\times 10^{-6}\text{old meter}1\; \backslash ,\; \backslash text\{new\; meter\}\; \backslash approx\; 1.533474\; \backslash times\; 10^\{-6\}\; \backslash ,\; \backslash text\{old\; meter\}$

• Second remains unchanged

Result

• New $hchc$ = 1

• New $GG$ = 1

Embedding Scaling Factors into Units

By embedding these scaling factors into the units of mass and length, we see that the gravitational force equation simplifies to:

$$F=\frac{m_1{m}_2}{r^2}F\; =\; \backslash frac\{m\_1\; m\_2\}\{r^2\}$$

where the units for mass ($\text{kg}\backslash text\{kg\}$) and length ($\text{m}\backslash text\{m\}$) have been redefined according to the scaling factors. This effectively removes the need for $GG$ as a separate constant, as its role is embedded in the scaled units.

Conclusion

By redefining the units of measurement, we demonstrate that the gravitational constant $G$ can be understood as a unit-scaling factor. This simplifies the gravitational equations and highlights the intrinsic relationships between mass, distance, and gravitational force. This perspective aligns with the use of natural units in theoretical physics and offers a unified framework for understanding physical laws.

Appendix A:

1. Original Gravitational Force Equation:

The gravitational force is given by:

$$F = G \frac{m_1 m_2}{r^2}$$

where:

• $G$ is the gravitational constant ($G = 6.674301 \times 10^{-11} \, \text{m}^3 / (\text{kg} \cdot \text{s}^2)$),
• $m_1$ and $m_2$ are the masses in kilograms,
• $r$ is the distance in meters.

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2. New Units (Scaled Kilogram and Meter):

You scale the kilogram ($\text{kg}$) and the meter ($\text{m}$) by the following factors:

$$1 \, \text{scaled kg} = k_m \, \text{original kg}, \quad k_m = \frac{1}{5.455511 \times 10^{-8}}$$ $$1 \, \text{scaled meter} = k_r \, \text{original meter}, \quad k_r = \frac{1}{1.533474 \times 10^{-6}}$$

Thus, the relationship between the new units and the original ones is:

$$m_1' = \frac{m_1}{k_m}, \quad m_2' = \frac{m_2}{k_m}, \quad r' = \frac{r}{k_r}$$

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3. Force in the New Units:

In the new system, $G$ is scaled to 1:

$$G' = 1 \, \text{scaled m}^3 / (\text{scaled kg} \cdot \text{s}^2)$$

The force equation becomes:

$$F' = G' \frac{m_1' m_2'}{r'^2}$$

Substituting the scaled quantities back into the equation:

$$F' = 1 \cdot \frac{\left(\frac{m_1}{k_m}\right) \left(\frac{m_2}{k_m}\right)}{\left(\frac{r}{k_r}\right)^2}$$

Simplify the fractions:

$$F' = \frac{m_1 m_2}{k_m^2} \cdot \frac{k_r^2}{r^2}$$

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4. Requiring $F' = F$:

For $F'$ to be equal to the original force $F$, we must have:

$$\frac{k_r^2}{k_m^2} = G$$

Using the scaling factors:

$$k_m = 1 / (5.455511 \times 10^{-8}), \quad k_r = 1 / (1.533474 \times 10^{-6})$$

Calculate $k_r^2 / k_m^2$:

$$k_r^2 = \left(\frac{1}{1.533474 \times 10^{-6}}\right)^2 = 4.246602 \times 10^{11}$$ $$k_m^2 = \left(\frac{1}{5.455511 \times 10^{-8}}\right)^2 = 3.363076 \times 10^{15}$$ $$\frac{k_r^2}{k_m^2} = \frac{4.246602 \times 10^{11}}{3.363076 \times 10^{15}} = 6.674301 \times 10^{-11} = G$$

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5. Conclusion:

The scaling factors for the meter ($k_r$) and kilogram ($k_m$) are chosen such that the ratio $k_r^2 / k_m^2$ equals $G$,ensuring that the force calculated using scaled units matches the force calculated in the original system.

This guarantees that:

$$F^{\mathrm{\prime }}=F$$

So yes, the scaling approach we have outlined does result in the correct force when using scaled units for mass and distance.

Appendix B:

Let's work through this to demonstrate how these definitions of h and G using alpha and beta generate the correct results.

First, let's define our constants:c = 299,792,458 m/s (speed of light)
h = 6.62607015 × 10^-34 J⋅s (Planck's constant)
G = 6.67430 × 10^-11 m^3 kg^-1 s^-2 (Gravitational constant)Now, let's solve for alpha and beta using the given equations:

1. h = (alpha^3 * beta) / c
2. G = (alpha^3 / beta)

Known Values:

• Meter scaling factor ($\alpha \backslash alpha$):

$$\alpha =1.53843951260968407858\times 10^{-6}\text{m}\backslash alpha\; =\; 1.53843951260968407858\; \backslash times\; 10^\{-6\}\; \backslash ,\; \backslash text\{m\}$$

• Kilogram scaling factor ($\beta \backslash beta$):

$$\beta =5.45551124829157414485\times 10^{-8}\text{kg}\backslash beta\; =\; 5.45551124829157414485\; \backslash times\; 10^\{-8\}\; \backslash ,\; \backslash text\{kg\}$$

Equations:

1. Planck's constant:

$$h=\frac{{\alpha }^3\cdot \beta }{c}h\; =\; \backslash frac\{\backslash alpha^3\; \backslash cdot\; \backslash beta\}\{c\}$$

2. Gravitational constant:

$$G=\frac{{\alpha }^3}{\beta }G\; =\; \backslash frac\{\backslash alpha^3\}\{\backslash beta\}$$

Solving for $hh$:

\[ h = \frac{(1.53843951260968407858 \times 10^{-6})3 \cdot 5.45551124829157414485 \times 10^{-8}}{299,792,458} \]

Calculations:

1. ${\alpha }^3\backslash alpha^3$:

\[ \alpha^3 = (1.53843951260968407858 \times 10^{-6})3 \]

$${\alpha }^3\approx 3.637060204\times 10^{-18}{\text{m}}^3\backslash alpha^3\; \backslash approx\; 3.637060204\; \backslash times\; 10^\{-18\}\; \backslash ,\; \backslash text\{m\}^3$$

2. $hh$:

$$h=\frac{3.637060204\times 10^{-18}\cdot 5.45551124829157414485\times 10^{-8}}{299,792,458}h\; =\; \backslash frac\{3.637060204\; \backslash times\; 10^\{-18\}\; \backslash cdot\; 5.45551124829157414485\; \backslash times\; 10^\{-8\}\}\{299,792,458\}$$

$$h\approx \frac{1.986451686\times 10^{-25}}{299,792,458}h\; \backslash approx\; \backslash frac\{1.986451686\; \backslash times\; 10^\{-25\}\}\{299,792,458\}$$

$$h\approx 6.62607015\times 10^{-34}\text{J}\cdot \text{s}h\; \backslash approx\; 6.62607015\; \backslash times\; 10^\{-34\}\; \backslash ,\; \backslash text\{J\}\; \backslash cdot\; \backslash text\{s\}$$

Solving for $GG$:

$$G=\frac{3.637060204\times 10^{-18}}{5.45551124829157414485\times 10^{-8}}G\; =\; \backslash frac\{3.637060204\; \backslash times\; 10^\{-18\}\}\{5.45551124829157414485\; \backslash times\; 10^\{-8\}\}$$

$$G\approx 6.67430\times 10^{-11}{\text{m}}^3{\text{kg}}^{-1}{\text{s}}^{-2}$$